Problem: Skipper Dan is sailing his ship through a swift current. Before entering the current, his ship was moving with a velocity (speed and direction) of $\vec{v_1}$. The direction of $\vec{v_1}$ is due west, and its speed is $20\,\text{km/h}$. Now that the ship is in the current, however, it is moving with a velocity $\vec{v_2}$. The direction of $\vec{v_2}$ is $40^\circ$ south of west, and its speed is $25\,\text{km/h}$. (Assume "due east" is $0^\circ$, "due north" is $90^\circ$, and so on.) What is the speed of the current?
Answer: Consider vector $\vec c$ (depicted below), which represents the current. We can imagine how it would cause Skipper Dan's boat to change direction and speed up a little bit. It is reasonable to assume that the velocity of the current added with the initial velocity of Skipper Dan's boat equals the resultant velocity of his boat. $ \vec c + \vec{v_1} = \vec{v_2}$ Before finding $\vec c$, we'll need to split both vectors into their components. $\vec{v_1}$ is $20\,\text{km/h}$ due west, which can be written as $(-20, 0)$. $\vec{v_2}$ can be broken into its components using trigonometry. We know that "due west" is $180^\circ$, so $40^\circ$ south of west must be $220^\circ$. Since we also know that the magnitude of $\vec{v_2}$ is $25\,\text{km/h}$, we can find the $x$ and $y$ components as follows: $x = 25\cos{(220^\circ)} \approx -19.151$ $y = 25\sin{(220^\circ)} = -16.070$ Therefore, $\vec{v_2} = (-19.151,\, -16.070).$ We can now solve for $\vec c$. $\begin{aligned} \vec c + \vec{v_1} &= \vec{v_2}\\\\ \vec c &= \vec{v_2} - \vec{v_1}\\\\ \vec c &= (-19.151,\, -16.070) - (-20, 0)\\\\ \vec c &= (0.849,\, -16.070) \end{aligned}$ We can find the magnitude of $\vec c$ (i.e., the speed of the current) using the Pythagorean theorem. $\begin{aligned} \| \vec c \|^2 &= 0.849^2 + (-16.070)^2\\\\ \| \vec c \| &= \sqrt{0.849^2 + (-16.070)^2}\\\\ \| \vec c \| &\approx 16.1 \text{ km}/\text{h} \end{aligned}$ $\vec c$ is pointing into the fourth quadrant, so we can find its direction (call it $\theta~$ ) using the arctangent function and adding $360^\circ$. (Figure not drawn to scale.) $\begin{aligned} \tan \theta &= \dfrac{-16.070}{0.849}\\\\ \theta &= \arctan{ \left ( \dfrac{-16.070}{0.849} \right ) } \\\\ \theta&\approx -87^\circ \end{aligned}$ Adding $360^\circ$ to this result gives us the actual direction, $273^\circ$. The speed of the current is $16.1 \text{ km}/\text{h}$. The current is flowing in a direction of $273^\circ$.